Let $x(t)$ a solution of $$x'=-x(t^2-x^2),$$ so that $|x(t_0)|\lt |t_0|$. Show that for all $|t|\gt |t_0|$, $|x(t)|\lt |t|$.

Suppose that $t_0\geq 0$. Consider

Then $(t_0,x(t_0))$ is over the red line. I don't see why: $x$ solution of the above ODE implies that the graph of $x$ stay between the blue lines to the right of the red line and to the left of the orange line.

The way the problem is currently stated, it is not true. I numerically solved the equation using $x(0)=.75$, and if we let $t_0=-1$, $|x(-1)|<1$. However, $|x(1)|>1$.
–
robjohn♦Apr 25 '12 at 23:19

@robjohn Thanks. I just copy the problem as is in the notes of the course. By now, I'll just skip this exercise.
–
leoApr 25 '12 at 23:34

If you take the problem as stated and restrict to either $t\ge0$ or $t\le0$, then it is true. It is only the correlation between $x(t_0)$ and $x(-t_0)$ that is bad. Are you studying flow?
–
robjohn♦Apr 25 '12 at 23:41

@robjohn, I'm studying ordinary differential equations. If you refer to this kind of flow then I'm not. In my attempt to solve the exercise, I put the restriction $t_0\geq 0$. Is that a good approach to the problem? what do you think?
–
leoApr 25 '12 at 23:50

that will give you a workable problem. However, you should also be able to solve the problem for $t_0<0$ as well. The problems are not exactly isomorphic.
–
robjohn♦Apr 26 '12 at 2:09

4 Answers
4

For simplicity, multiply the equation by $2x$ and subtract $2t$ to get
$$
(x^2-t^2)'=2x^2(x^2-t^2)-2t\tag{1}
$$
First, let's deal with $t>0$.

$(1)$ says that if $x^2-t^2<0$, then we also have $(x^2-t^2)'<0$. Therefore, if $x(t_0)^2-t_0^2<0$, we have $x(t)^2-t^2<0$ for all $t\ge t_0$.

Next, let's deal with $t<0$.

$(1)$ says that if $x^2-t^2\ge0$, then we also have $(x^2-t^2)'\ge0$. Therefore, if $x(t)^2-t^2\ge0$, we have $x(t_0)^2-t_0^2\ge0$ for all $t_0\ge t$. The contrapositive says that for all $t\le t_0$, if $x(t_0)^2-t_0^2<0$, then we have $x(t)^2-t^2<0$.

A Graphical Explanation

Consider the direction field (aka slope field and flow diagram) for the solutions. That is, the vector field $\color{#0000c0}{\frac{\mathrm{d}}{\mathrm{d}t}(t,x)}$ which is tangent to all solutions $\color{#c00000}{(t,x)}$.

$\hspace{3cm}$

The important part to consider is the flow across the boundary $x^2=t^2$ between the regions where $\color{#00c000}{x^2< t^2}$ and $\color{#c000c0}{x^2>t^2}$. Notice that the on the boundary, the flow is horizontal since $x'=x(x^2-t^2)=0$.

$\hspace{4.2cm}$

Thus, when $t<0$ (on the left), a solution can only move from the region where $\color{#00c000}{x^2< t^2}$ to the region where $\color{#c000c0}{x^2>t^2}$. When $t>0$ (on the right), a solution can only move from the region where $\color{#c000c0}{x^2>t^2}$ to the region where $\color{#00c000}{x^2< t^2}$.

Suppose that the set $$C:=\{t\gt t_0:x(t)\geq t\}$$ is not empty, thus, since $C$ is bounded below by $t_0$, $$\tau=\inf C$$ is well defined. We can proof (by taking a sequence $C\ni t_n\to\tau$ and assuming continuity of $x$) that $\tau\in C$. In particular $\tau\gt t_0$ and then $$x(\tau)\geq\tau\gt 0,$$ then, there exist an $\alpha\gt 0$ such that $x$ is positive over $]\tau-\alpha,\tau+\alpha[\subset]t_0,\infty[$.

Then for all $t\in]\tau-\alpha,\tau[$, by the definition of $\tau$, $t\notin C$ (otherwise $t\in C$ with $t\lt\inf C$), so $0\lt x(t)\lt t$ and then $$x'(t)=x(t)(x(t)^2-t^2)\lt 0,$$
so $x$ is decreasing over $]\tau-\alpha,\tau[$.

If $x(t_0)=0$ then $x(t)=0$ for all $t\ge t_0$. Without loss of geneality, we assume $x(t_0)\ne0$.

If $0&ltx(t_0)&ltt_0$, you want to show that $0&ltx(t)\le t$ for all $t\ge t_0$. Now, $x'(t_0)=-x(t_0)(t_0^2-x(t_0)^2)&lt0$. By uniqueness of solutions $x(t)>0$ for all $t\>t_0$. And since the right hand side of the equation is negative on the region $\{(t,x):0&ltx&ltt,t>t_0\}$, it is easy to see that $x$ is decreasing.

If $x(t_0)=t_0$ then $x'(t_0)=0$. Taking derivatives in both sides of the equation we get $x''(t_0)=-2\,t_0\,x(t_0)&lt0$. Thus $x'(t)&lt0$ on some interval $(t_0,t_0+\delta)$, $\delta>0$, $x(t)&ltt$ on that same interval, and the argument in the previos paragraph carries through.

@leo: note that $$ \frac{\mathrm{d}}{\mathrm{d}t}\log|x|=x^2-t^2 $$ When $|x|$ is close to $0$, $\log|x|$ is close to $-\infty$, yet the slope of $\log|x|$ is finite. Even if the slope were positive, which it isn't, it would take a long time to climb to something not close to $-\infty$. The closer $x$ gets to $0$, the longer it would take. Unfortunately, the slope of $\log|x|$ is not positive.
–
robjohn♦Apr 25 '12 at 22:48

$x(t)=0$ is certainly a solution. Uniqueness of solution (the right hand side is $C^\infty$, so it is locally Lipschitz,) implies that if the solution vanishes at some point, it is identically equal to $0$.
–
Julián AguirreApr 26 '12 at 8:22